Question: Solve the equation. $\dfrac{dy}{dx}=-\dfrac{x+6}{7y^2}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\pm\sqrt{-\dfrac{x+6}{7}+C}$ (Choice B) B $y=C\sqrt{-\dfrac{x+6}{7}}$ (Choice C) C $y=C\sqrt[3]{-\dfrac{3x^2}{14}-\dfrac{18x}{7}}$ (Choice D) D $y=\sqrt[3]{-\dfrac{3x^2}{14}-\dfrac{18x}{7}+C}$
Solution: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{x+6}{7y^2} \\\\ -7y^2\,dy&=(x+6)\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} -7y^2\,dy&=(x+6)\,dx \\\\ \int -7y^2\,dy&=\int (x+6)\,dx \\\\ -\dfrac{7y^3}{3}&=\dfrac{x^2}{2}+6x+C_1 \\\\ y^3&=-\dfrac{3x^2}{14}-\dfrac{18x}{7}+C \\\\ y&=\sqrt[3]{-\dfrac{3x^2}{14}-\dfrac{18x}{7}+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\sqrt[3]{-\dfrac{3x^2}{14}-\dfrac{18x}{7}+C}$